Let y²=12x be the parabola and S be its focus. Let PQ be a focal chord of the parabola such that (SP

Q: Why is the focal distance $SP = a + x_P$?

This is a fundamental definition of a parabola: the distance from any point to the focus is equal to the distance to the directrix. Since the directrix is $x = -a$, the distance is $x_P - (-a) = x_P + a$. It is a beautiful geometric shortcut that saves you from using the distance formula with square roots.

Q: Why does $t_P \cdot t_Q = -1$ for a focal chord?

The slope of the line passing through $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$ is $m = \frac{2a(t_2 - t_1)}{a(t_2^2 - t_1^2)} = \frac{2}{t_1 + t_2}$. If the line passes through the focus $(a, 0)$, the slope is also $\frac{2at_1 - 0}{at_1^2 - a} = \frac{2t_1}{t_1^2 - 1}$. Equating these leads to $t_1 t_2 = -1$. It is a classic property that simplifies many focal chord problems.

Q: Can we use the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ instead of the diameter form?

Yes, but it is much more tedious. You would need to find the midpoint as the center and the distance between $P$ and $Q$ as the diameter. The diameter form $(x - x_P)(x - x_Q) + (y - y_P)(y - y_Q) = 0$ is much more elegant and less prone to calculation errors.

Question

Let $y^{2}=12x$ be the parabola and S be its focus. Let PQ be a focal chord of the parabola such that $(SP)(SQ)=\frac{147}{4}.$ Let C be the circle described taking PQ as a diameter. If the equation of a circle C is $64x^{2}+64y^{2}-\alpha x-64\sqrt{3}y=\beta.$ then $\beta-\alpha$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at the parabola $y^2 = 12x$. It is not just a curve; it is a path, a reflection of light, a trajectory.

By comparing $y^2 = 12x$ with the standard form $y^2 = 4ax$, we immediately identify $4a = 12$, which means $a = 3$. The focus $S$ is therefore at $(3, 0)$. This is our anchor point.

### The Parametric Dance

Now, consider a focal chord $PQ$. A focal chord is a line segment passing through the focus. To handle this algebraically, we use parametric coordinates.

Let $P = (3t^2, 6t)$. Because $PQ$ is a focal chord, the product of the parameters of its endpoints must be $-1$. This is a powerful property.

If $t_P = t$, then $t_Q = -1/t$. Thus, $Q = (3/t^2, -6/t)$. This substitution transforms a complex geometric problem into a manageable algebraic one.

### The Focal Distance Insight

The problem gives us the product $(SP)(SQ) = \frac{147}{4}$. We recall the focal distance property: for any point on a parabola, the distance to the focus is $a + x_P$.

So, $SP = 3 + 3t^2 = 3(1 + t^2)$ and $SQ = 3 + 3/t^2 = 3(1 + 1/t^2)$. When we multiply these, we get:

$$9(1 + t^2)(1 + 1/t^2) = 9(1 + 1/t^2 + t^2 + 1) = 9(t^2 + 1/t^2 + 2) = 9(t + 1/t)^2$$

This is the moment of clarity. We have reduced the entire expression to a simple square.

### Solving for the Parameters

Equating $9(t + 1/t)^2 = \frac{147}{4}$, we find:

$$(t + 1/t)^2 = \frac{147}{36} = \frac{49}{12}$$

Taking the square root, $t + 1/t = \frac{7}{2\sqrt{3}}$. This leads to the quadratic $2\sqrt{3}t^2 - 7t + 2\sqrt{3} = 0$.

Factoring this, we find $t = \frac{\sqrt{3}}{2}$ or $t = \frac{2}{\sqrt{3}}$. These are the parameters for our endpoints $P$ and $Q$. Substituting $t = \frac{2}{\sqrt{3}}$ into our parametric forms, we get $P = (4, 4\sqrt{3})$ and $Q = (9/4, -3\sqrt{3})$.

### Constructing the Circle

We are tasked with finding the equation of a circle with $PQ$ as a diameter. The diameter form $(x - x_P)(x - x_Q) + (y - y_P)(y - y_Q) = 0$ is our best friend here.

Substituting our coordinates, we get:

$$(x - 4)(x - 9/4) + (y - 4\sqrt{3})(y + 3\sqrt{3}) = 0$$

Expanding this, we get $x^2 - (4 + 9/4)x + 9 + y^2 - (4\sqrt{3} - 3\sqrt{3})y - 36 = 0$, which simplifies to:

$$x^2 + y^2 - \frac{25}{4}x - \sqrt{3}y - 27 = 0$$

### Final Calculation

To match the given form $64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta$, we multiply the simplified equation by $64$:

$$64x^2 + 64y^2 - 400x - 64\sqrt{3}y - 1728 = 0$$

Thus, $\alpha = 400$ and $\beta = 1728$. The final answer is:

$$\beta - \alpha = 1728 - 400 = **1328**$$

Enter Numerical Value:

Visualized Solution

Parabola and Focus

Parametric Coordinates of Focal Chord

Focal Distances $S P$ and $S Q$

Product $(S P) (S Q)$

Solving for $t + \frac{1}{t}$

Solving the Quadratic for $t$

Coordinates of $P$ and $Q$

Circle Equation in Diameter Form

Expanding the Circle Equation

Scaling the Equation

Comparing to find $α$ and $β$

Final Calculation of $β - α$

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Parametric Dance

The Focal Distance Insight

Solving for the Parameters

Constructing the Circle

Final Calculation

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